In my lecture on stochastic processes it is stated that the natural filtration $\mathcal{F_t}^0=\sigma(\forall s\leq t: \omega\mapsto \omega(s)$ is measurable$)$ is not a good choice for Brownian motion because for example $$ L=\lim_{t\searrow 0}\frac{B_t}{\sqrt{2t\log|\log t|}} $$ is not measurable w.r.t. $\mathcal{F}^0$ but w.r.t. $\mathcal{F_t}=\bigcap_{s>t}\mathcal{F_s}^0$.

Why is this the case? I have no clue why $L$ is $\mathcal{F}$-measurable and why not $\mathcal{F}^0$-measurable and it's nowhere explained.

Can someone of you help me please?


  • $\begingroup$ The second filtration is right continuous $\endgroup$ – badatmath Jan 7 at 16:00
  • $\begingroup$ The natural filtration is such that $\mathcal{F_0^0}=\{\phi, \Omega\}$ because it's trivial sigma algebra (a $B_0=0$), so if $L$ is not constant you are done. Using time inversion you can look at the law of iterated logarithm to get a non constant result for L unless mistaken. $\endgroup$ – TheBridge Jan 7 at 16:56

Notice that for each $\epsilon>0$, $$ L=\inf_{0< s<\epsilon, s\in\Bbb Q}\sup_{t\in(0,s), t\in\Bbb Q}{B_t\over \sqrt{2t\log|\log t|}} $$ is $\mathcal F^0_\epsilon$ measurable. (I have replaced the limit in your definition of $L$ with limit superior; the latter always exists, the former doesn't exist, for a.e. Brownian path.) It follows that $L$ is $\mathcal F_0$ measurable. On the other hand, $\mathcal F^0_0=\sigma\{B_0\}$ and $L$ is clearly not a function of $B_0$ alone.


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